$\sum\limits_{n=1}^{\infty } \dfrac{3(x+1)^n}{n\cdot4^n} $ What is the interval of convergence of the series? Choose 1 answer: Choose 1 answer: (Choice A) A $-5 \le x \le 3$ (Choice B) B $-4 \le x<2$ (Choice C) C $-5 \le x<3$ (Choice D) D $-4 \le x \le 2$
We use the ratio test. For $x\neq-1$, let $a_n=\dfrac{3(x+1)^n}{n\cdot4^n}$. $\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|= \left| \dfrac{x+1}{4}\right|$ The series converges when $\left| \dfrac{x+1}{4}\right|<1$, which is when $-5<x<3$. Now let's check the endpoints, $x=-5$ and $x=3$. Letting $x=-5$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{3(-5+1)^n}{n\cdot4^n} &= \sum\limits_{n=1}^{\infty } \dfrac{3(-4)^n}{n\cdot4^n} \\\\ &= \sum\limits_{n=1}^{\infty } \dfrac{3(-1)^n(4)^n}{n\cdot4^n} \\\\ &=3 \sum\limits_{n=1}^{\infty }\dfrac{(-1)^n}{n} \end{aligned}$ This is a multiple of the alternating harmonic series, which is known to converge. Letting $x=3$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{3(3+1)^n}{n\cdot4^n} &= \sum\limits_{n=1}^{\infty } \dfrac{3(4)^n}{n\cdot4^n} \\\\ &=\sum\limits_{n=1}^{\infty }\dfrac{3}{n} \\\\ &=3 \sum\limits_{n=1}^{\infty }\dfrac{1}{n} \end{aligned}$ This is a multiple of the harmonic series, which is known to diverge. In conclusion, the interval of convergence is $-5 \le x<3$.